Corrections and additions for the book Perturbation Analysis of Optimization Problems by J.F. Bonnans and A. Shapiro. Version of March 28, 2013

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1 Corrections and additions for the book Perturbation Analysis of Optimization Problems by J.F. Bonnans and A. Shapiro Version of March 28, 2013 Some typos in the book that we noticed are of trivial nature and do not need an explanation. There are, however, more subtle corrections that need to be made. There are also simple extensions and additions to the material presented in the book which are worthwhile to mention. 1 Corrections Section In Section 2.1.4, X and X are assumed to be paired locally convex topological vector spaces equipped with respective compatible topologies. The closure operation in that section is taken with respect to these topologies. In particular, if X is a nonreflexive Banach space and X is its dual, then the standard pair of compatible topologies is the strong (norm) topology in X and weak topology in X (see the discussion after Definition 2.26 on page 25). In that case the closure in the right hand sides of formulas (2.32) and (2.33), while in the space X, should be taken in the weak topology. This may lead to a confusion since in the Basic Notation cl( ) is defined with respect to the norm (strong) topology. If X is a reflexive Banach space, then one can use strong topologies in X and X. Page 52 In formula (2.115) the inequality h(ω) 0 should be replaced by h(ω) 0. Also, in the following line x(ω) > 0 should be replaced by x(ω) < 0. Page 66, Proposition For the given definition of the set T formula (2.171) is in error. Correct definition of the set T should be T := {h X : G(x 0 ) + DG(x 0 )h K}. (1.1) In order to see why (2.171) is wrong let us consider the convex set K Y. Let y 0 K and T K (y 0 ) be the corresponding tangent cone. Then K y 0 + T K (y 0 ) and hence dist(d, T K (y 0 )) = 0 for any d K y 0. Formula (2.172) then follows, under Robinson s constraint qualification, by Stability Theorem However, if Y is infinite dimensional, it is not true in general that dist(y, K) = o( y y 0 ), y y 0 + T K (y 0 ). (1.2) 1

2 Consider, for example, space Y := l 1 and the set K := { (y n ) l 1 : y 1 ny 2 n, n = 2,... }. (1.3) This is a nonempty closed convex set and T K (0) = {(y n ) l 1 : y 1 0}. (1.4) We have here that for every n-th coordinate vector e n and t > 0, te n T K (0), and for n (2t) 1, dist(te n, K) = inf y [0,t] {ny2 + t y} = t + (4n) 1 t = te n. (1.5) Consequently, (1.2) does not hold at y 0 = 0. The assertion of Corollary 2.91 is correct. The fact that, under Robinson s constraint qualification, T i Φ(x 0 ) = T Φ (x 0 ) and equation (2.173) holds can be easily derived from Stability Theorem 2.87 and the relation that for any fixed d T K (y 0 ) and y(t) = y 0 + td + o(t) it follows that dist(y(t), K) = o(t), t > 0. Page 68, proof of Prop Error and correction kindly indicated to us by Nguyen Huy Chieu. The multifunction M(x, t) defined at the beginning of the proof is not closed as is wrongly stated. It should be redefined as follows, denoting by B Y the closed unit ball of Y : { DG(x0 )x t((k G(x M(x, t) := 0 )) B Y ), if t 0,, if t < 0. Page 82. Formula (2.230) of Proposition is correct. However, its proof is imprecise. First line of the proof of Proposition should be replaced by the following. It follows from (2.229), applied to f, that x f (x) iff f (x) = x, x f (x ). (1.6) Since, by the Fenchel-Moreau-Rockafellar Theorem 2.113, f = f the above equation is equivalent to (2.231). Starting from equation (2.231) the proof can be completed as in the book. Page 215. In the right hand side of formula (3.191), F G(x 0 )w should be replaced by DF (x 0 )w. Page 242. Line before equation (3.268), for any z Z should be replaced by for any z C. 2

3 Page 270. Errors kindly indicated to us by Alexey F. Izmailov. In the first line of (4.21), read D x g i (x 0, u 0 ) (and not Dg i (x 0, u 0 )). In the second line of (4.22), read εd instead of d. Page 275. Formula (4.39) should be cl conv ( x S(u0 )D u f(x, u 0 ) ). (1.7) Under the assumptions of theorem 4.13, the set x S(u0 )D u f(x, u 0 ) indeed is compact. Therefore, in case the space U is finite dimensional, the convex hull of that set is also compact and hence is closed, and hence the topological closure in the above formula can be omitted. Page 302. Line before equation (4.131), h n := x n x 0 should be replaced by h n := κ 1 n (x n x 0 ). Pages 404. In equation (5.11): read F (x 0, u 0 ) instead of D x F (x 0, u 0 ). Page 420. Error kindly indicated to us by Diethard Klatte and Ebrahim Sarabi. Theorem 5.20 is correct, but the proof of the converse part, starting p. 420 on line 14, must be corrected in the following way. We first recall a classical property of strong regularity, and then state the corrected proof. A classical property of strong regularity Strong regularity (SR) for the inclusion ϕ(z) + N(z) 0 (1.8) is defined as the local existence of a univalued, Lipschitz inverse for the perturbed linearized system ϕ(z 0 ) + Dϕ(z 0 )(z z 0 ) + N(z) δ. (1.9) We claim that SR holds iff the local existence of a univalued, Lipschitz inverse holds for the perturbed system ϕ(z) + N(z) δ. (1.10) Indeed, suppose that the SR holds for the linearized system (1.9). It is shown in Theorem 5.13 that this implies that the system (1.10) has a local solution. Proof of this is based on a fixed point theorem (contraction principle). Converse implication of existence of local solution can be proved in a similar way. As far as Lipschitz continuity is concerned we can proceed as follows. We may by a translation argument assume that z 0 = 0 and that ϕ(z 0 ) = 0. We must analyze the relation Az + N(z) δ, (1.11) 3

4 where A := Dϕ(z 0 ). We check first the Lipschitz property. Consider sequences δ k and δ k and z k, z k, all converging to 0, such that Az k + N(z k ) δ k and Az k + N(z k) δ k. We may write these relations as ϕ(z k ) + N(z k ) δ k + ϕ(z k ) Az k, and similarly for z k, so that by our hypothesis z k z k = O( δ k δ k + ϕ(z k) Az k ϕ(z k ) Az k ). (1.12) Since the r.h.s. is of order O( δ k δ k ) + o( z k z k ), the Lipschitz property follows. Uniqueness of the solution follows whenever δ is close enough to 0. Proof of the converse part of thm 5.20 (starting p. 420 on line 14). From the uniform second order growth condition, and since DG(x 0 ) is onto and hence Robinson s constraint qualification holds, theorem 5.17 implies that there exist neighborhoods V of 0 X Y and V X of x 0 such that for all δ V, the canonically perturbed problem Min x f(x) δ 1, x ; G(x) + δ 2 K. (1.13) has a locally unique solution x(δ), and since DG(x 0 ) is onto, there exists a unique µ(δ) Y such that (x(δ), µ(δ)) is solution of the canonically perturbed optimality system, i.e. Df(x) + DG(x) µ = δ 1 ; G(x) + δ 2 K. (1.14) It remains to show that the mapping (x( ), µ( )) is Lipschitz continuous. So consider two elements ˆδ and δ in V, and the associated solutions of (1.14) denoted by (ˆx, ˆµ) and ( x, µ), respectively. Since DG(x 0 ) is onto, we may apply the stability theorem 2.87 to the relation G(x) + δ 2 = 0. It follows that (reducing the size of the neighborhoods if necessary) there exists x X such that G(x) + ˆδ 2 = G( x) + δ 2 ; x x = O( ˆδ 2 δ 2 ). (1.15) Let δ 1 := Df(x) + DG(x) µ. Then (x, µ) is a solution of (1.14) for δ = (δ 1, ˆδ 2 ). We have that δ 1 δ 1 = Df(x) Df( x) + (DG(x) DG( x)) µ. This and (1.15) imply δ 1 δ 1 = O( ˆδ δ ). By proposition 4.32 we have that x ˆx = O( ˆδ δ ), and so by (1.15) again, x ˆx = O( ˆδ δ ). Since DG(x 0 ) is onto, there exists ε > 0 such that, for any x close enough to x 0, we have that DG(x ) µ ε µ. Since Df(ˆx) + G(ˆx) ˆµ = 0, we deduce that ˆµ ε 1 Df(ˆx) = O(1). Since Df(ˆx) Df( x) = G( x) µ G(ˆx) ˆµ + δ 1 ˆδ 1 = DG( x) ( µ ˆµ) + (DG( x) DG(ˆx)) ˆµ + δ 1 ˆδ 1. (1.16) 4

5 We deduce that µ ˆµ = O( x ˆx + ˆδ δ ) = O( ˆδ δ ). The conclusion follows. Pages In the proof of theorem 5.60, some terms are missing in the optimality system at the end of page 460 and for the quadratic program at the top of page 461. The missing terms are D 2 g i (x 0, u 0 )((h 1, u 1 ), (h 1, u 1 )) in the expansion of constraints, and (D 2 (x,u)x G(x 0, u 0 )(h 1, u 1 )) λ 1 in the expansion of D x L, as well as the corresponding λ 1 D 2 (x,u)x G(x 0, u 0 )((h 1, u 1 ), h 2 ) in the cost function of the quadratic program. One should read, starting at the display at the bottom of page 460: D 2 (x,u)xl(x 0, λ 0, u 0 )(h 2, u 2 ) + D 3 (x,u)(x,u)xl(x 0, λ 0, u 0 )((h 1, u 1 ), (h 1, u 1 )) +(D 2 (x,u)xg(x 0, u 0 )(h 1, u 1 )) λ 1 + DG(x 0, u 0 ) λ 2 = 0, D 2 g i (x 0, u 0 )((h 1, u 1 ), (h 1, u 1 )) + Dg i (x 0, u 0 )(h 2, u 2 ) = 0, i {1,..., q} I 1 +, D 2 g i (x 0, u 0 )((h 1, u 1 ), (h 1, u 1 )) + Dg i (x 0, u 0 )(h 2, u 2 ) 0, i I u1 (x 0, u 0, h 1 ) \ I 1 +, λ 2i Dg i (x 0, u 0 )(h 2, u 2 ) = 0, i I u1 (x 0, u 0, h 1 ) \ I 1 +. The above system has a unique solution, since it is the optimality system of the quadratic problem whose objective, etc. Min h2 D 3 (x,u)(x,u)(x,u)l(x 0, λ 0, u 0 )((h 1, u 1 ), (h 1, u 1 ), (h 2, u 2 )) +D 2 (x,u)(x,u)l(x 0, λ, u 0 )((h 2, u 2 ), (h 2, u 2 )) +λ 1 D 2 (x,u)xg(x 0, u 0 )((h 1, u 1 ), h 2 ) s.t. Dg i (x 0, u 0 )(h 2, u 2 ) = 0, i {1,..., q} I 1 +, Dg i (x 0, u 0 )(h 2, u 2 ) 0, i I u1 (x 0, u 0, h 1 ) \ I 1 +, Page 476. Second line after equation (5.171), C -smooth and G xw r should be replaced by C -smooth and G W r. Page 516. First line. The sentence: If x 0 is a stationary point of (P ), then the first order growth condition holds at x 0 iff C(x 0 ) = {0}, should be replaced by: If x 0 is a stationary point of (P ) and the extended MF constraint qualification holds, then the first order growth condition holds at x 0 iff C(x 0 ) = {0}. 5

6 2 Additional material Page 301. The result of Proposition 4.52 can be extended as follows. Proposition 2.1 Suppose that the assumptions of proposition 4.52 hold and let (ˆx(u), ˆλ(u)) be a stationary point of (P u ) such that ˆx(u) x 0 as u u 0. Then ˆx(u) x 0 = O ( u u 0 1/2). (2.17) Proof. It suffices to show that for any sequence u n converging to u 0 and x n := ˆx(u n ), it follows that κ n = O ( ) τn 1/2, where κn := x n x 0 and τ n := u u 0. Denote also λ n := ˆλ(u n ). By passing to a subsequence if necessary we can assume that λ n converges to some λ 0 Λ(x 0 ). We have that f(x n, u n ) = L(x n, λ n, u n ) and f(x 0, u 0 ) = L(x 0, λ 0, u 0 ). Because of Robinson s constraint qualification, there exist points x n which are feasible for the unperturbed problem (P u0 ) and such that x n x n = O(τ n ). Consequently by the second order growth condition we obtain f(x n, u n ) f(x 0, u 0 ) = f(x n, u 0 ) f(x 0, u 0 ) + f(x n, u 0 ) f(x n, u 0 ) + f(x n, u n ) f(x n, u 0 ) c x n x 0 2 c 0 τ n cκ 2 n c 1 τ n c 1 τ n κ n, where c, c 0 and c 1 are some constants with the constant c being positive. Since it follows that L(x 0, λ 0, u n ) L(x 0, λ 0, u 0 ) = O(τ n ), L(x n, λ n, u n ) L(x 0, λ 0, u n ) cκ 2 n c 2 τ n (1 + κ n ) (2.18) for some constant c 2. We also have that L(x n, λ n, u n ) L(x n, λ 0, u n ) = λ n λ 0, G(x n, u n ) and λ n, G(x n, u n ) = 0. Moreover, since K is generalized polyhedral we can assume by passing to a subsequence if necessary that λ 0, G(x n, u n ) = 0. Together with (2.18) this implies L(x n, λ 0, u n ) L(x 0, λ 0, u n ) cκ 2 n c 2 τ n (1 + κ n ). (2.19) Consider the mapping F (z, u) := (D x L(x, λ, u), G(x, u)), where z := (x, λ), and let z n := (x n, λ n ) and z 0 := (x 0, λ 0 ). Since the multifunction Γ(z) is monotone we obtain by the generalized equations (4.115) that z n z 0, F (z n, u n ) F (z 0, u 0 ) 0. (2.20) 6

7 On the other hand z n z 0, F (z n, u n ) F (z 0, u 0 ) = x n x 0, D x L(x n, λ n, u n ) D x L(x 0, λ 0, u 0 ) λ n λ 0, G(x n, u n ) G(x 0, u 0 ) = x n x 0, D x L(x n, λ 0, u n ) D x L(x 0, λ 0, u 0 ) λ n λ 0, G(x n, u n ) G(x 0, u 0 ) D x G(x n, u n )(x n x 0 ) = κ 2 nd 2 xxl(x 0, λ 0, u n )(h n, h n ) + c 3 κ n τ n + o(κ 2 n) 1 2 λ n λ 0, D 2 xxg(x 0, u 0 )(h n, h n ) + O(τ n ) = κ 2 nd 2 xxl(x 0, λ 0, u n )(h n, h n ) + c 3 κ n τ n + o(κ 2 n) + o(τ n ) for h n := (x n x 0 )/κ n and some constant c 3. Together with (2.18) and (2.20) this implies that 0 cκ 2 n + c 4 κ n τ n + c 5 τ n for some constants c, c 4 and c 5 with the constant c being positive. It follows then that κ n = O ( ) τn 1/2, which completes the proof. 7